Question: The lifespans of zebras in a particular zoo are normally distributed. The average zebra lives $27.3$ years; the standard deviation is $6$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a zebra living between $15.3$ and $45.3$ years.
Answer: $27.3$ $21.3$ $33.3$ $15.3$ $39.3$ $9.3$ $45.3$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $27.3$ years. We know the standard deviation is $6$ years, so one standard deviation below the mean is $21.3$ years and one standard deviation above the mean is $33.3$ years. Two standard deviations below the mean is $15.3$ years and two standard deviations above the mean is $39.3$ years. Three standard deviations below the mean is $9.3$ years and three standard deviations above the mean is $45.3$ years. We are interested in the probability of a zebra living between $15.3$ and $45.3$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the zebras will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the zebras will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of zebras between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular zebra living between $15.3$ and $45.3$ years is ${95\%} + \color{orange}{2.35\%}$, or $97.35\%$.